What is the solution for y^{''}-3y^'=x+3 ?

The solution for y^{''}-3y^'=x+3 is y=c_{1}e^{3x}+c_{2}-(x^2)/6-(10x)/9

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y^{''}-3y^'=x+3

y^{^{''}} - 3 y^{^{'}} = x + 3

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{6} - \frac{10 x}{9}

Solution steps

y^{''} (x) - 3 y^{'} (x) = x + 3

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{6} - \frac{10 x}{9}

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{6} - \frac{10 x}{9}

x^{^{''}} - x^{^{'}} - 2 = 5 cos (2 t)

(d^{2} - 8 d + 20) \cdot y = 100 x^{2} - 26 x \cdot e^{x}

(d^{2} + 4 d + 5) \cdot y = e^{- 2 x}

(d^{2} + 1) y = sec (x)

(d^{2} - 4) y = x^{2} e^{3} x

What is the solution for y^{''}-3y^'=x+3 ?
The solution for y^{''}-3y^'=x+3 is y=c_{1}e^{3x}+c_{2}-(x^2)/6-(10x)/9